CPhill

Let the top of the statue  be point A

Let the bottom be point B

Place  the mirror between Kiran and the statue such that from  Kiran's point of view, the whole statue shows in the mirror....call this  point, C

Call the top of Kiran's head point D

Call the place where he stands  point E

Because  the angle of  incidence of light  = its angle of reflection, then angle ACB   =Angle DCE

And Angle  ABC  =  Angle DEC  = right angles

So..........by AA congruency.....triangle  ABC  is similar to triangle DEC

Here's the idea   {using a palm tree rather than a statue ]  : 

Now....note something else.....if the mirror is placed  305 ft from the statue = BC   and Kiran stands exactly 5.8 ft from the mirror = EC....we have this relationship

AB / BC  =   DE / EC

305 / 305  =  5.8  / 5.8

Which is true

Furthermore

The tangent of angle ACB [ angle of incidence]  =   AB / BC  = 305/305  =  1

And

The tangent of  angle DCE  [angle of reflection ]  = 5.8 / 5.8  = 1

So....since the tangents are equal.....then  we know that the angle of  incidence  = angle of reflection

The only way that we can get two  positive roots is if we have something like this :

(x  - a)^2   =   0             take both roots

x  - a   =  0    [  both roots of   0   are  -0   and  + 0   =  0   ]

x  = a           (assuming that  a  >  0  )

2^(x + y)  * 3^(x - y)  * 6^(2x + 2y)   =  72

2^(x + y) *  3^(x - y) * ( 6^2)^(x + y)  =  72

2^(x + y) * 3(x - y) * 36(^(x + y)  =  72

( 2 * 36)^(x + y)  * 3^(x - y)  =  72

(72)^(x + y)  * 3^(x - y)  =  72

Since

72^(1) * 3^(0)  =   72

This will be true when

x +  y   =   1

x -  y  =     0              add these

2x   =  1

x   =1/2

And  

1/2 + y   = 1

y  = 1/2

(x,y)   =  (1/2, 1/2)

Thanks, EP......here's a general way of solving something like this

Assuming that  the road is uphill one way and downhill the other way....we have that....

Avg rate    =    Total Distance  / Total Time

Call the total  distance  =  2D

The total time is   ( D/ 30)  + (D /20) 

So...we have that

Avg rate  =     ( 2D)                               2D                              2* 600

                  _____________  =       ______________  =       ______     =  24 mph

                     D/30 + D/24                D [ 30 + 20] /600                50

CORRECTED   !!!!

We can use  the triangle  inequality to solve this

Let   S  be the unknown  side

So....

S + 5  >  7           S + 7  >   5             7  +   5   >   S

S  >  2                  S  >   -2                   12  >    S

                                                              S   <  12

Taking the most restrictive positive interval for S......  

S  = ( 2 , 12)

We have   9  integers  that  will satisfy this   (3, 4, 5, 6, 7, 8 , 9 , 10 , 11)

So....the probability  =     9  / 15    =    3   /  5

225 / 36     has two   roots  : (-15/6)     and  (15/6)   

If we square both of these.....we get   225 /36

But.....since    "a"  is positive....we need   15 / 6    =   5   / 2

Does that make sense  ???

Determine the set of all real x satisfying (x^2+3x-1)^2<9
Enter your answer in interval notation.

Take both roots and we have that

x^2 + 3x - 1   =    3                     x^2   + 3x - 1  = - 3  

x^2  + 3x  - 4   = 0                     x^2  +  3x  + 2   =   0

(x + 4) (x - 1)  = 0                     (x  + 2)  ( x + 1)   =  0

Setting each factor to 0 and solving for x  gives the answers that  x  = { -4, - 2,- 1, 1 }

So....we have the possible answers  

(-inf, -4)    ( -4, -2)   ( -2, -1)   (-1, 1)    or  ( 1 inf )

Testing a point in each interval in the original inequality we see that the  solution intervals are

(-4, -2)  U   (-1, 1)

ab  =  11.25       ⇒  b  =  11.25 / a       (1)

ac  = 20   ⇒  c  = 20  / a        (2)

bc  = 36         (3)

Sub (1)  and (2)   into (3)

(11.25 /a ) (20 / a)  = 36

225/ a^2  =  36       rearrange as

a^2  =  225/36       take the positive root

a  =  15 / 6    =   5 / 2

So....using (2).....

c =  20 / ( 5/2)   =   8

Lay out a  rectangle with  a width of 5  and a height of  3

As shown here :  https://www.desmos.com/calculator/qnelrpyogs

The ant will cross 7  distinct squares

So....in a  rectangle with a width of 20  and a height of 12......the ant will cross 4 times as many squares  = 28

Sequences with one T  =

(TAR)  (TAG) (TAE)

(TRG) (TRE)

(TGE)

And each of these can be arranged in 6 ways....so....6 * 6  =   36       (!)

Sequences with two T's  =

(TTA) (TTR) (TTG) (TTE)

And each of these can be arranged in 3 ways ,,,,so   4 * 3  =  12     (2)

So...adding (1)  and (2)   we get  48 arrangements with at least one T